2 KClO3 ----> 2 KCl + 3 O2
Suppose the "inert" material in the mixture was not really inert, but that is reacted with oxygen from the atmosphere when heat was applied. Would the apparent mass of oxygen liberated be larger or smaller than it should be?
Then, suppose that some sample material was lost by spilling during heating. Would the experimentally determined percentage KClO3 be larger or smaller than the true value?
If the inert material soaked up some of the liberated oxygen, wouldn't you think less free O2 would be liberated?
If you spill some of the sample the mass O2 liberated will be less so the percent will be less.
To answer the first question, let's analyze the reaction equation:
2 KClO3 → 2 KCl + 3 O2
According to the stoichiometry of the balanced equation, for every 2 moles of KClO3 that react, 3 moles of O2 are liberated. Therefore, if the "inert" material in the mixture reacted with oxygen from the atmosphere when heat was applied, it would lead to an increased amount of available oxygen. This means that the apparent mass of oxygen liberated would be larger than it should be.
Now, let's address the second question regarding the loss of sample material during heating.
If some of the sample material is lost by spilling during heating, it would result in a lower amount of KClO3 available for the reaction. Since the experimentally determined percentage of KClO3 is calculated based on the mass of KClO3 used in the reaction, the lost sample would lead to a lower mass of KClO3 in the calculation. As a result, the experimentally determined percentage of KClO3 would be smaller than the true value.
It is worth noting that in experimental settings, it is important to handle and measure substances accurately to minimize errors and obtain more reliable results.